If velocity of a charged particle is doubled and strength of magnetic field is halved, then radius becomes :

Question

If velocity of a charged particle is doubled and strength of magnetic field is halved, then radius becomes : (A) 8 times (B) 2 times (C) 4 times (D) 3 times

Tardigrade Admin · Accepted Answer

Force =q v B=(m v2/r)=( centripetal force) ⇒ r=(m v/q B) ∴ r prime=(m × 2 v/q × (B)2)=4 r

Q. If velocity of a charged particle is doubled and strength of magnetic field is halved, then radius becomes :

8 times

2 times

4 times

3 times

Force $= q v B = \frac{m v ^{2}}{r} = ($ centripetal force)
$\Rightarrow r = \frac{m v}{qB}$
$∴ r^{'} = \frac{m \times 2 v}{q \times \frac{B}{2}} = 4 r$

Q. If velocity of a charged particle is doubled and strength of magnetic field is halved, then radius becomes :

8 times

2 times

4 times

3 times

Force =qvB=rmv2​=( centripetal force) ⇒r=qBmv​ ∴r′=q×2B​m×2v​=4r

Force $= q v B = \frac{m v ^{2}}{r} = ($ centripetal force)
$\Rightarrow r = \frac{m v}{qB}$
$∴ r^{'} = \frac{m \times 2 v}{q \times \frac{B}{2}} = 4 r$